Final answer:
When genes A and B are 15 map units apart and cross AaBb with aabb, approximately 85% of the offspring are expected to be dominant for both traits A and B, accounting for the genetic linkage and recombination frequency.
Step-by-step explanation:
The question inquires about the expected proportion of offspring that are dominant for both traits A and B when genes A and B are 15 map units apart and the dihybrid AaBb is mated with a tester aabb. Given that there is a specified map unit distance between the genes, this indicates that the genes are linked, which affects the assortment of alleles during meiosis and hence the probability of certain genotypes in the offspring.
Considering the distance between the traits is 15 map units, crossing over will occur between these two genes during meiosis in some of the gametes, but not all. Typically, 15 map units correspond to a 15% recombination frequency, which means 15% of gametes will be recombinant, and 85% will be non-recombinant or parental types. Therefore, when crossing AaBb with aabb, the expected proportion of offspring that are dominant for both traits (which requires A and B alleles from the dihybrid) will primarily include parental types as the majority.
The calculation for the desired genotype (AB) is the product of the probabilities of obtaining a dominant allele A (since aa is not possible in this case, all A alleles contribute to the phenotype) and a dominant allele B from the AaBb parent, which is (1 - recombination frequency for AB) or (1 - 0.15). So, the expected proportion is approximately 85% or 0.85.